Question

Find the zeroes of the quadratic polynomials given below. Find the sum and product
of the zeroes and verify relationship to the coefficients ofterms in the polynomial.
(i) p(x) = x2 - x - 6 (ii) p(x) = x2 - 4x + 3
(iii) p(x) = x2 - 4
(iv) p(x) = x2 + 2x + 1​

Answer 1

Answer:

First I have find the zeroes and then I have find the relationship between co-efficients.

Step-by-step explanation:

Hope it helps.

Answer 2

Solution :

We know that standard quadratic polynomial ax² + bx +c

let the zeros are α and β .

So, α + β = -b/a

αβ = c/a

(i) p(x) = x² – x – 6

x² – 3x + 2x – 6 =0

x(x-3) + 2 ( x-3) =0

(x- 3)( x+2) =0

x = 3 and x =-2

From coefficients α + β = -b/a = 1 ------ eq 1

from obtained zeros 3-2 = 1 --------- eq 2

sum of zeros are equal ;eq 1 and eq2 are equal.

αβ = c/a = -6

3(-2) = -6

(ii) p(x) = x² – 4x + 3

x² – 3x - x + 3 =0

x( x-3) -1 ( x- 3) =0

(x - 3)( x -1) =0

x = 3 and x = 1

α + β = -b/a = 4

3+ 1 = 4

αβ = c/a = 3

3(1) = 3

(iii) p(x) = x² – 4

x² – 4 =0

(x+2)( x- 2) =0

x = -2 and x = 2

α + β = -b/a = 0

-2 + 2 = 0

αβ = c/a = -4

2(-2) = -4

(iv) p(x) = x² + 2x + 1

x² + x + x + 1 =0

x ( x+1) +1 (x +1) =0

(x +1) (x+1) =0

x = -1 and x = -1

α + β = -b/a = -2

-1 -1 = -2

Pls mark me brain list answer